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If y = \(\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}}\), find \(\rm \frac{{dy}}{{dx}}\)
1. \(\rm\frac{1}{2}\left( {secx.cotx + cosecx.tanx} \right)\)
2. \(\rm \frac{1}{2}\left( {secx.cotx - cosecx.tanx} \right)\)
3. \(\rm \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)
4. \(\rm \frac{1}{2}\left( {secx.tanx - cosecx.cotx} \right)\)

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Correct Answer - Option 3 : \(\rm \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)

Concept used:

Trigonometry formula

sin 2x = 2sin x cos x


y = \(\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}}\)

y = \(\rm \frac{{\left( {sinx - cosx} \right)}}{{2.sinx.cosx}}\)

⇒ y = \(\rm \frac{{sinx}}{{2.sinx.cosx}} - \;\frac{{cosx}}{{2.sinx.cosx}}\)

⇒ y = \(\rm \left( {\frac{1}{{2cosx}}} \right)\; - \left( {\frac{1}{{2sinx}}} \right)\)

⇒ y = \(\frac{1}{2}\) (secx - cosecx)

Differentiate both sides, we get

⇒ \(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{{d\left( {secx} \right)}}{{dx}} - \frac{{d\left( {cosecx} \right)}}{{dx}}} \right]\)

⇒ \(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)

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